Chapter 18C:
Alternating Current
Contents:
7. Alternating Current
18-7 Alternating Current
Current from a battery flows steadily
in one direction (direct current, DC).
Current from the electricity supply
varies sinusoidally (alternating
current, AC), reversing in direction.
Voltage varies sinusoidally:
where 𝑉0 = peak voltage (V)
𝜔 = angular frequency (s−1)
𝜔 = 2𝜋𝑓
where 𝑓 = frequency (Hz) so
𝑉 = 𝑉0 sin 2𝜋𝑓𝑡
𝑉 = 𝑉0 sin 𝜔𝑡
Alternating Current
𝑉 = 𝑉0 sin 2𝜋𝑓𝑡
where 𝑉0 = peak voltage.
The 'peak-to-peak' voltage
𝑉𝑝𝑝 = 𝑉0 − −𝑉0 = 2𝑉0
Frequency 𝑓 is number of
complete oscillations per second
and period 𝑇 is time for one complete oscillation
1
𝜔
2𝜋
𝑓=
; 𝑓=
; 𝑇=
𝑇
2𝜋
𝜔
e.g. Mains frequency in SA is 50 Hz so period of waveform is
1
1
𝑇= =
= 0.02 s = 20 × 10−3 s = 20 ms
𝑓 50
Alternating Current
If voltage V across resistor,
current in resistor given by
"Ohm's law" for DC and AC:
𝑉 𝑉0
𝐼 = = sin 𝜔𝑡 = 𝐼0 sin 𝜔𝑡
𝑅 𝑅
where 𝐼0 = 𝑉0Τ𝑅 = peak current
Note if 𝑉 = 𝑉0 sin 2𝜋𝑓𝑡
then 𝐼 = 𝐼0 sin 2𝜋𝑓𝑡
- the two sinusoidal*
waveforms are "in phase"
(the maxima and minima
coincide with each other)
*sinusoidal: sin or cos function
Alternating Current
Because 𝑉 and 𝐼 go equally
positive and negative, the
average value of each is zero.
But power 𝑃 = 𝐼𝑉 = 𝐼2 𝑅 so
𝑃 = 𝐼02 𝑅 sin2 2𝜋𝑓𝑡
Now sin2 varies between 0 and 1
with average value 1Τ2 
The average power is then
1 2
1 2
ത
ത
𝑃 = 𝐼0 𝑅 or 𝑃 = 𝑉0 Τ𝑅
2
2
(since also 𝑃 = 𝑉 2 Τ𝑅)
or 𝑃ത = 𝐼ഥ2 𝑅 (𝐼ഥ2 = average of 𝐼2 )
Alternating Current
If we now take the square root of 𝐼ഥ2 (= 𝐼02 Τ2) we get the
rms (root-mean-square) current, similarly for rms voltage:
𝐼rms = 𝐼ഥ2 =
𝑉rms = 𝑉2 =
𝐼rms and 𝑉rms are sometimes
called "effective" values of
current and voltage because
they can be substituted into
the (dc) formulae for power:
2 𝑅 etc
𝑃ത = 𝐼rms 𝑉rms = 𝐼rms
𝐼0
2
𝑉0
2
= 0.707 𝐼0
= 0.707 𝑉0
Alternating Current
Normally, when we specify an ac voltage we give the rms
value, e.g. "mains voltage in South Africa is 230 V" means
this is the rms value 𝑉rms = 𝑉0Τ 2 . The peak voltage is
𝑉0 = 2 𝑉rms = 2 230 = 325 V
Example 18-13 (modified):
Calculate the resistance and peak
current in a 1500 W hairdryer
connected to a 230 V ac line.
𝑃ത
1500
𝐼rms =
=
= 6.5 A
𝑉rms
230
𝐼0 = 2 𝐼rms = 2 6.5 = 9.2 A
𝑉rms 230
𝑅=
=
= 35.4 Ω
𝐼rms
6.5